Since is monotone and , we have iff , i.e., . Does this help?
Hello,
I have the following Problem:
A={(x,y) element of R2}: [e^(3-x^2-(y-2)^2)<=1)]
Is A convex? Is it open? Is it bounded?
I don`t know how to draw this set, the only thing I know that the definition of a convex set is that it is possible to join every element of the set with another element within the set with a straight line, without leaving the set.
I know that this set is not open, because it does not include all of its boundary points (because its domain has the sign <=)
I don`t know if the set is bounded, I only know that the definition of bounded is, that the set could be inside of a giant circle.
Sorry for my bad English. I hope someone can give me a good explaination about how to solve problems like this.
It would help if you point out exactly what you don't understand. The function is increasing. We have for x < 0, and for x > 0. Therefore, is equivalent to , i.e., . Now, the locus of is a circle with center (0, 2) and radius . Obviously, consists of points that are farther from (0, 2) then .
I gave this example another thought.
Could I solve it in another way too? For example:
A={(x,y) element of R2}: [e^{(3-x^2-(y-2)^2)}<=1)]
Than I have e^{x} is <=1 for x <=0
Next I test the function 3-x^2-(y-2)^2<=0 for convexity, so I calculate the Hessian which is:
H_{f}=-4
The Hessian is negative so it`s convex.
e^{x} is convex but it's domain is concave, so the set is concave.
Is this definition true?
This is not a function, this is an inequality. Which function do you mean?
Can you explain how you did this?
The domain of e^{x} is the set of real numbers, which is convex. I am not even sure what a concave set is. What theorem did you use here?
A definition cannot be true or false.
I understood that my answer was completely stupid. I was thinking totally in a wrong direction.
Now another question, if we turn the set around and say:
A={(x,y) element of R2}: [e^{(3-x^2-(y-2)^2)}>=1)]
Than I start with:
e^{x} >=0
therefore
(3-x^2-(y-2)^2) >= 0
this is true for:
y=-((3-x^{2})^{1/2}-1) or y=(3-x^{2})^{1/2}+1
so this function is:
-closed because it's complement is open
-bounded because it can be drawn in a large enough circle
-convex because every point inside the set can be connected to each point within the set with a straight line, without leaving the set.
Is this true?
Than I finally understood the problem. If it's not true I have to push my head against the wall :-D